+0  
 
0
34
1
avatar+817 

Let c be a real number, and consider the system of quadratic equations
y = x^2 - 9x + c,
y = 8x^2 - 13x.
For which values of c does this system have:

(a) Exactly one real solution (x,y)?

(b) More than one real solution?

(c) No real solutions?

Solutions to the quadratics are (x,y) pairs.

 Nov 20, 2023
 #1
avatar+36 
0

Hint 1: Use the quadratic formula, most notably, the discrimant.

 

Hint 2: Part a) is a single number, the rest are inequalities.

 

 

 

 

 

 

 

 

 

 

 

Here's the answer for part (a):

If you put the 2 equations together, you will get \(x^2-9x+c=8x^2-13x\), from which you will get \(7x^2-4x-c=0\). In order for their to be only 1 solution, the discriminant must be equal to \(0\). The discrimant of any quadratic is \(b^2-4ac\) (It is the part in the square root of the quadratic formula, which is the formula that pops up whenever you press the LaTeX button on your screen.)

 

\(b^2-4ac = 0\), so

\(16+28c=0\), from which you can solve

\(c=-\frac{16}{28}=-\frac{4}{7}\)

 Nov 22, 2023

1 Online Users

avatar